Math, asked by sumitpanchal142007, 10 months ago

A pedestal is constructed in the form of the frustum of a
pyramid, the sides of the square ends of the frustum being
360 cm and 160 cm and its slant height 260 cm. Find volume,
lateral surface area including area of the top and cost of
construction @ 50/cubic meter and plastering it @ 10/
metre square.​

Answers

Answered by Anonymous
0

The cost of construction is Rs 1012.80.

Slant height = 260 cm (Given)

Square ends of frustum = 360 cm and 160 cm (Given)

Base = (360 - 160)/2

= 100 cm

Slant height = a² + b² = c²

= a²  + 100² = 260²

= a²  = 57600

= a = √57600

= a = 240 cm

Therefore, height is 240 cm

Volume = h/3 ( Area 1 + Area 2 + √ Area 1 x Area 2

= 2.4/3 ( 3.6 x 3.6 + 1.6 x 1.6 + √(3.6 x 1.6) )

= 0.8 ( 17.92) = 14.336 m³

Therefore, the volume is 14.336 m³

Lateral surface area -

For one face = 1/2 × (3.6 + 1.6) x 2.6 = 6.76 m²

For four faces = 6.76 x 4 = 27.04 m²

Surface area needed to plaster:

=  1.6 x 1.6 + 27.04

= 29.6 m²

Thus, the total surface area is 42.56 m²

Cost of construction - 1 m³ = Rs 50

14.336 m³ = 50 x  14.33

= Rs 716.80

The cost of construction is Rs 716.80

Cost of plastering -

1 m² = Rs 10

Thus, for 29.6 = 29.6 x 10 = Rs 296

The cost of plastering is Rs 296

Total cost: = 716.80 + 296

= Rs 1012.80

Similar questions